Application of Definite Integral I been solving this problem for an hour, but I cant get the correct answer. Please help me with this.
 A: If you split the desired violet area into $2$ equals parts by drawing the line $x=\frac32$, you can evaluate it. In fact you only need to solve the system below
\begin{cases}
\int_0 ^\frac32 \sqrt{4-x^2} = A+S\\
\int_0 ^2 \sqrt{4-x^2} = A+2S\\
\end{cases}
letting $S=$ half of the violet area, and $A =$ the sector of the first circle obtained by removing the  entire violet area. By solving the system you get
\begin{cases}
\left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^\frac32 = A+S\\
\left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^2 = A+2S\\
\end{cases}
\begin{cases}
\frac 34\sqrt{4-\frac{9}{16}}+2\arcsin\left(\frac 34 \right)= A+S\\
2\arcsin\left(1\right)= A+2S\\
\end{cases}
\begin{cases}
\frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)= A+S\\
\pi = A+2S\\
\end{cases}
$$\Rightarrow S=A+2S-(A+S)=\pi-\left(\frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)\right)\approx 0.0549\Rightarrow \\\text{Violet Coloured Area}\approx 2\cdot 0.0549 \approx 0.109$$
